Beam span and loading.
Free Body Diagram
The first step is to draw a free body diagram (see Figure T301.1) that graphically illustrates the beam and the external forces acting on it.
Figure T301.1 T1  Free Body Diagram.
Static Equilibrium.
The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:
 @ A  Vertical reaction, R_{y}, and @ B  Horizontal reaction, R_{y}
 @ B  Vertical reaction, R_{y}
For the beam, this gives a total of three unknown forces, for which there are three equations of equilibrium:
Therefore: The beam can be said to be statically determinate.
Horizontal Reactions
The beam, under the loading indicated above, is not subjected to a lateral load, i.e. there is no load acting horizontally either to the left or the right. Therefore to maintain static equilibrium, where the sum of the horizontal forces equates to zero there cannot be any horizontal action at either A or B. Therefore:
Vertical Reactions
For a simply supported beam, where the only load is a universally distributed load (a constant load along the full length of the beam), the vertical reactions at the supports can be calculated one of two ways.
 By taking moments about one support to determine the other, or
 By simply recognising that the load will be equally distributed between the two vertical supports.
Consider the reaction at "B". This can be calculated by taking moments about "A".
Figure T301.1 illustrates the concept of "positive" and "negative" rotation of a load or reaction when taking moments about A. The UD load acts in a clockwise direction about A and is considered "POSITIVE" ("+ve"), whereas the vertical reaction Vb is acting in an upwards direction tending to rotate anticlockwise about A and is therefore considered to be "NEGATIVE" ("ve").
Figure T301.1 T1  Positive & Negative Rotation About "A"
For the reaction at B, using the third equation of equilibrium and taking moments about A, we can say:
We now formulate an expression that equates the moments generated by the UD load (w) and the vertical reaction (Vb), when acting about A, to zero:
Substituting in the tutorial data gives:
Finally, solving for the reaction R_{B}:
For the vertical reaction at A, we adopt the first equation of equilibrium:
Equating the vertical actions (loads and reactions) to zero, we can say:
NOTE: The direction of the arrows indicating the load / reactions. Vertically upwards = POSITIVE, vertically downward = NEGATIVE.
Substituting in the tutorial data and the calculated value of the reaction at B, gives:
Finally, solving for the reaction R_{A}:
The method above outlines the way in which to calculate the reactions from first principles. For the more general case, where there is a universally distributed load (UDL) across the full span of the beam the vertical reaction at either A or B is simply a case of the total load divided by 2, i.e:
Shear Force
A Shear Force Diagram is a means of graphically illustrating the change in shear force along the length of the beam. The horizontal axis represents the length of the beam, and the vertical axis represents the shear force.
The first step in constructing a shear force diagram is to draw a horizontal line that represents the beam. It is then a case of plotting the reaction at A and then incrementally along the beam plotting the change in shear force  Figure 301.2 illustrates this principle.
Figure T301.2 Construction of the Shear Force Diagram
Figure 301.1 (i) to (vi) shows the incremental construction of the Shear Force Diagram.
 (i)  A vertical line is drawn that represents the reaction at A. If the reaction acts up, then the line representing this is drawn up. NOTE: The lines making up the shear force diagram are drawn in the same direction as the actions they represent.
 (ii)  @ 1m along the beam we consider everything to the left of the 1m line. From 0m to 1m the only change in the actions is 1m worth of the UD load w, acting down. Therefore the value of the shear force at 1m = Ra  w.
 This same principle is carried on for the full length of the beam. Therefore @ 5m (v)  along the beam we consider everything to the left of the 5m line. From 0m to 5m the only change in the actions is 5m worth of the UD load w, acting down. Therefore the value of the shear force at 5m = Ra  5w.
 At the right hand end of the beam (vi) the shear force at 6m = Ra  6w. At this point the vertical (upwards) reaction Rb acts. We therefore draw a vertical line representing Rb. As this is the end of the beam, and to maintain static equilibrium, the shear force at 6m (Ra  6w) should equal the vertical reaction Rb, such that:
Solution Shear Force Diagram
Bending Moment
The Bending Moment Diagram is a graph that shows the variation in bending along a beam. For its construction we must consider "HOGGING" and "SAGGING" in a beam and how these produce either "POSITIVE" or "NEGATIVE" bending moments.
NOTE: The calculations associated with bending moments have nothing to do with the direction of rotation, only if the applied action results in positive or negative bending.
Figure T301.3 Positive and Negative Bending Moments
The maximum bending moment, for a beam with a universally distributed load along its full length, is at the midspan point.
It is also useful to remember that where the shear force is zero coincides with the maximum bending in the beam.
The position of the maximum bending moment (X), measured from A:
The procedure for taking moments about the centre (or any point on a beam) is as follows:
Step 1


"CUT": The beam at the point about which you are taking moments. In this case we cut the beam in the center where the bending moment is the maximum. 
Step 2


"BIN": Ignore one half of the beam and fix the half you are interested in. In this case we are looking at the left hand end. This section has been given a notional fixed support. "ONE END IN": Working from the end of the beam to the fixed support, consider each action in turn and formulate expressions for the moments they generate. 
Step 3


Reaction Ra: The vertically upwards reaction at A causes TENSION in the bottom of the beam. It is therefore a SAGGING moment and POSITIVE. Taking moments about the fixed end gives:

Step 4


UDL  w: The UDL acts vertically downwards causing TENSION in the top of the beam. It is therefore a HOGGING moment and NEGATIVE. Taking moments about the fixed end gives:

A useful mnemonic to remember: "CUT, BIN, ONE END IN" 
Putting the above two expressions together gives a single expression for the bending moment at the midspan of the beam.
Substituting in the design data:
Solving, gives the final midspan moment:
For the general case where a simply supported beam has a UDL across the full span the bending moment can be calculated as follows (you should remember this!):
Bending Moment Diagram
It can be shown that the shape of the bending moment diagram for a simply supported beam with a UD load across the full length is that of a parabola:
Figure T301.4
Consider the beam portion shown in Figure T301.4.
If we take moments about an arbitrary point, X from the support at A, we can derive an expression for the bending moment at that point.
"CUT": Cut the beam at X.
"BIN": Ignore anything to the right of X.
"ONE END IN": Working from the end of the beam, derive expressions for the bending moments.
For the reaction at A (SAGGING):
For the UD load (HOGGING):
Combining these gives an equation for the Bending Moment at X and also the equation of a parabola.
Solution Bending Moment Diagram
NOTE: When construction the bending moment diagram:
 The bending moment is ALWAYS drawn on the tension side of the beam.
 POSITIVE, SAGGING bending moments where the bottom of the beam is in tension are drawn below the line.
 NEGATIVE, HOGGING bending moments where the top of the beam is in tension are drawn above the line.